Question

Before the development of reliable batteries, miners’ lamps burned acetylene produced by the reaction of calcium carbide with water. A typical lamp used 0.75 L of acetylene per hour at
1.00 atm pressure and 20°C. How many grams of calcium carbide and how many grams of water had to be in the lamp for a 4-hour shift?

CaC₂(s) + H₂O(l) -> C₂H₂(g) + CaO(s)

Answer 1

MW of C₂H2 = 26. At 1.00 atm pressure and 20° e 22.4 L C₂ H2 = 26 g C₂ H2 0.75 L C2 H₂ ² 26 X 0.75 - 22.4 = 0.871 g C2 H2 The lamp used acetylene in one hown = 0.8719 Themefore the lamp used acetylene in four houn = (0.87 X 4) g = 3.484 g

CaC2 (8) + H₂O (1) C2 H2 (g) + Cao (5) MW = 64.0999 MW = 18 g 26 g 26 g C₂ H2 produce from = 64.099 g Cala 3.484 g C2 H2 produce from = 64.09983.484 = 8.589 g Cacz 26 64.099 g Calz reacts with 18 g H2O to produce one mole of C₂H2. Therefore, 8.598 g Cal, reacts with 18x8.598 64.099 = 2.414 g H2O g H2O Thus 8.589 g calcium carbide and 2.414g of H2O had to be in the lamp for a 4-hour shift.